What is the domain of f(x)= x/(2x^2-5)-sqrt(x) ?

The domain of f(x)= x/(2x^2-5)-sqrt(x) is 0 sqrt(5/2)

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domain of f(x)= x/(2x^2-5)-sqrt(x)

domain

f (x) = \frac{x}{2 x ^{2} - 5} - x

0 \leq x < \frac{5}{2} or x > \frac{5}{2}

Interval Notation

[0, \frac{5}{2}) \cup (\frac{5}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{5}{2}, x = - \frac{5}{2}

Combine real regions and undefined points for final function domain

0 \leq x < \frac{5}{2} or x > \frac{5}{2}

- 3 cos (5 x)

\frac{x ^{2} + 4 x + 7}{x + 3}

f (x) = 7 - x + 5

x + 2 y = 6

f (x) = \frac{x - 4}{x + 4}

What is the domain of f(x)= x/(2x^2-5)-sqrt(x) ?
The domain of f(x)= x/(2x^2-5)-sqrt(x) is 0<= x<sqrt(5/2)\lor x>sqrt(5/2)